My Math Forum The shape of a parabola

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 May 10th, 2012, 02:43 PM #1 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 The shape of a parabola I've been trying to wrap my head around why all quadratic equation parabolas with a leading coefficient of 1 (or just have the same leading coefficient) have the exact same shape, just a different location around the Cartesian graph. ex X^2 , X^2-2x, and X^2-10x All the parabolas will have the exact same shape, and increase the exact same amount from their respective vertexs- 0,0 1,-1 1,1 2, 0 increase of 1 2,4 3,3 of 3 3,9 4,8 of 5 It seems almost counter-intuitive that the graphs will always have the exact same shape, and move away from the vertex's at the exact same pace. I mean, there must be some pattern that when you are at the point where you increase or decrease X by any amount, Y will go up (plain English- the vertex) then it will precisely follow the parent graph X^2. I would like to visualize or grasp some concept of why this is. My squared drawings of dots were less than helpful.
 May 10th, 2012, 03:00 PM #2 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: The shape of a parabola Complete the squares and everything will be perfectly clear
May 10th, 2012, 03:15 PM   #3
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Re: The shape of a parabola

Quote:
 Originally Posted by fredde_fisk Complete the squares and everything will be perfectly clear
So basically think of it like
x^2-2x
(x-1)^2= 1
(x-1)^2-1=x^2-2x vs x^2
Okay so the graph will be the exact same, except in this case x will have to be 1 more to the right to compensate for the -1, and every single value of Y will be 1 less because of the constant. So now it is quite obvious the graph will behave in the exact same way.

Thanks for my noob reminder!

December 24th, 2012, 09:22 PM   #4
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Quote:
 Originally Posted by maxgeo All the parabolas will have the exact same shape
That depends on how you define "shape". Do all circles have the same shape, despite the fact that, by definition, they have different curvature if they are of different size?

December 28th, 2012, 06:22 PM   #5
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Re:

Quote:
Originally Posted by skipjack
Quote:
 Originally Posted by maxgeo All the parabolas will have the exact same shape
That depends on how you define "shape". Do all circles have the same shape, despite the fact that, by definition, they have different curvature if they are of different size?
Noob question: doesn't 'same shape' mean 'similar'?

 December 30th, 2012, 08:35 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,941 Thanks: 2210 Usually, yes, but the original poster seemed to have congruence in mind.
December 30th, 2012, 09:59 PM   #7
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Re: The shape of a parabola

Hello, maxgeo!

Did you understand fredde_fisk's suggestion?

Quote:
 I've been trying to wrap my head around why all quadratic equations with a leading coefficient of 1 (or just have the same leading coefficient) have the exact same shape, just a different location around the Cartesian graph.

Complete the square:

$\text{Given: }\:y \;=\;ax^2\,+\,bx\,+\,c$

[color=beige]. . . . . . . . [/color]$=\;a\left(x^2\,+\,\frac{b}{a}x\,+\,\frac{c}{a}\rig ht)$

[color=beige]. . . . . . . .[/color]$=\;a\left(x^2\,+\,\frac{b}{a}x\,+\,\frac{b^2}{4a^2 }\,+\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}\right)$

[color=beige]. . . . . . . .[/color]$=\;a\left(x^2\,+\,\frac{b}{a}x\,+\,\frac{b^2}{4a^2 }\right)\,+\,c \,-\,\frac{b^2}{4a}$

[color=beige]. . . . . . . .[/color]$=\;a\left(x\,+\,\frac{b}{2a}\right)^2\,+\,\left(\f rac{4ac\,-\,b^2}{4a}\right)$

$\text{This is the same graph as }y \,=\,ax^2$
[color=beige]. . . . . [/color]$\text{with the vertex moved to: }\,\left(-\frac{b}{2a},\;\frac{4ac-b^2}{4a}\right)$

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